Question

The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire and passing through the centre of mass of the system :

• A. $M{r}^2(1-\frac{4}{{\pi }^2})$
• B. $M{r}^2(1+\frac{4}{{\pi }^2})$
• C. $M{r}^2(1-\frac{{\pi }^2}{4})$
• D. $M{r}^2(1+\frac{{\pi }^2}{4})$

Answer 1

Answer: (A)

$M{r}^2(1-\frac{4}{{\pi }^2})$

$I_{semicircle}=\frac{1}{2}{I}_{circle}=\frac{1}{2}{M}_{circle}{r}^2=M_{semicircle}{r}^2$
But, $I_{semicircle}=I_{cm}+M_{semicircle}(\frac{2r}{\pi }{)}^2$ where $\frac{2r}{\pi }$ is the distance of the com from the centre of the semicircle.

$\therefore I_{cm}=I_{semicircle}-M_{semicircle}(\frac{2R}{\pi }{)}^2=M_{semicircle}{r}^2-M_{semicircle}(\frac{2r}{\pi }{)}^2=M_{semicircle}{r}^2(1-\frac{4}{{\pi }^2})$